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viva [34]
3 years ago
12

Michael is the captain of his school’s soccer team. What skill does Michael exhibit when he decides which player will take the p

enalty kick?
A. conflict resolution
B. sportsmanship
C. leadership
D. teamwork
Physics
1 answer:
aleksley [76]3 years ago
5 0
A because if he was a leader he’d take the kick himself and they’re are not all working to kick the ball so it’s not teamwork and it’s not sportsmanship because he isn’t stopping a fight or fixing something between two teams
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3 years ago
Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri
irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

d=\frac{2g(m1-m2)}{k}

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

E_{i}=E_{f}

K_{i}+U_{i}=K_{f}+U_{f} (1)

With K_{i} the initial kinetic energy, U_{i} the initial potential energy, K_{f} the final kinetic energy and U_{f} the final potential energy. Note that initialy the masses are at rest so K_{i} = 0, when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so K_{f} =0. So, equation (1) becomes:

U_{i}=U_{f} (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so U_{i}=U_{gi} and at final moment we have potential gravitational energy and potential elastic energy so U_{f}=U_{gf}+U_{ef}, using this on (2)

U_{gi}=U_{gf}+U_{ef} (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), U_{gi}=0 and we have by (3) :

0= U_{gf}+U_{ef} (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

0=-m1gd+m2gd+\frac{kd^{2}}{2}

dividing both sides by d

0=-m1g+m2g+\frac{kd}{2}

g(m1-m2)= \frac{kd}{2}

d=\frac{2g(m1-m2)}{k}, with k the constant of the spring and g the gravitational acceleration.

7 0
3 years ago
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Answer:

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Explanation:

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